{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solutionset7

# 344 from bottom the hoop then position and velocity

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: sses move along a (θ + α)so = speed of each + α),˙ (which you could also circle, the − dt2 2 3R 2 obtain by calculating x + y ). The Lagrangian is therefore ˙ ˙ √ d2 2g 2) (θ − αα2 + mgR cos(θ −(α) + cos(θ + α) =− θ − α). (348) L = 2(m/2)R ˙ dt2 R = mR2 α2 + 2mgR cos θ cos α. ˙ The normal coordinates are therefore (341) √ The equation of motion is θ + α = A1 cos(ω1 t + φ1 ), 2g where ω1 = , 3R g cos θ g cos θ 2 2mR α = −2mgR cos θ sin α =⇒ α ≈ − ¨ ¨ α =⇒ √2g ω= . (342) R R θ − α = A2 cos(ω2 t + φ2 ), where ω2 = . (349) R ◦ Note that if θ = 0, then ω = g /R, as it should. And if θ = 90 , then ω = 0, which Adding and subtracting these yields the angles, which when written in vector notamakes sense. tion are 6.44. Oscillating hoop with a pendulum θ 1 1 = B1 cos(ω2 t + φ2 ), (350) The positions of the masses cos(ω1 t + φ1 ) + B2 are   α 1 −1       ◦ ◦ (x, y ’s = R − A’s. If B θ = 0, we have θ) (343) where the B)1 are h...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online